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Basile [38]
2 years ago
11

A box is at rest on a slope

Physics
1 answer:
matrenka [14]2 years ago
6 0

Answer:

170 N

Explanation:

The following data were obtained from the question:

Mass (m) = 20 Kg

Angle (θ) = 30°

Acceleration due to gravity (g) = 9.8 m/s²

Weight (W) =?

The perpendicular component of the weight can be obtained by using the following formula:

W = mg × Cosθ

W = 20 × 9.8 × Cos 30

W = 196 × 0.866

W = 169.7 ≈ 170 N

Therefore, perpendicular component of the weight is 170 N

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Andrew pokes a marble, and the marble rolls down a ramp. The marble moves with speed. Which forces are acting on the marble in t
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The forces are Andrew poking the marble and then gravity pulling the marble downward
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What is the likely identity of a metal if a sample has a mass of 63.5 g when measured in air and an apparent mass of 60.2 g when
Gre4nikov [31]

Answer:

Gold

Explanation:

Given:

Mass of sample = 63.5 g

Mass of water = 60.2 g

Find:

Object

Computation:

Mass of water displaced = 63.5 g - 60.2 g

Mass of water displaced = 3.3 g

So, volume in water = 3.3 cm³

Density = Mass / Volume

Density = 63.5 g / 3.3

Density = 19.24

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7 0
2 years ago
An insect 5.00 mm tall is placed 20.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ
jenyasd209 [6]

Answer:

a) i = -9.63 cm ,    h ’= .0.24075 cm   erect

b)  i = 259.74 cm ,

Explanation:

For this exercise let's start by finding the focal length of the lens

               1 / f = (n-1) (1 / R₁ - 1 / R₂)

                1 / f = (1.70 -1)) 1 / ∞ - 1/13)

                1 / f = 0.0538

                 f = - 18.57 cm

Now we can use the constructor equation

             1 / f = 1 / o + 1 / i

             1 / i = 1 / f - 1 / o

              1 / i = -1 / 18.57 -1/20

               1 / i = -0.1038 cm

               I = -9.63 cm

For the height of the

image let's use magnification

                 m = h '/ h = - i / o

                  h ’= -h i / o

                  h ’= - 0.5 (-9.63) / 20

                  h ’= .0.24075 cm

b) we invert the lens

The focal length is

             1 / f = (1.70 -1) (1/13 - 1 / int)

              1 / f = 0.0538

             f = 18.57 cm

             1 / i = 1 / f -1 / o

             1 / I = 1 / 18.57 - 1/20

             1 / I = 3.85 10-3

             i = 259.74 cm

     

            h ’= - 0.5 259.74 / 20

             h ’= 6.4935 cm

7 0
3 years ago
The radius of curvature of a spherical mirror is 20cm.What is its focal length?
hichkok12 [17]

Answer:

\boxed{\sf Focal \ length = 10 \ cm}

Given:

Radius of curvature (R) of a spherical mirror = 20

To Find:

Focal length (f)

Explanation:

Formula:

\boxed{ \bold{\sf Focal \ length \ (f) = \frac{Radius \ of \ curvature \ (R)}{2}}}

Substituting value of R in the equation:

\sf \implies f =  \frac{20}{2}

\sf \implies f = \frac{ \cancel{2} \times 10}{ \cancel{2}}

\sf \implies f = 10 \: cm

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Figure 1 shows a wave movement during one second. What is the frequency of the wave
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This is 2 hertz.  You can mark out 2 full wavelengths in the second of time.
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